13th International Conference on Fracture June 16–21, 2013, Beijing, China -3- Figure 2. The considered single lap joint configuration 2.2. Linear elasticity solution for the single lap joint To obtain an efficient failure model a closed-form solution of the mechanical behavior of the single lap joint has been chosen in this work. Since the pioneering works by Volkersen [17] and the work by Goland and Reissner [8] many linear-elasticity solutions for the single lap joint have been developed and proposed in literature. Previous works by the authors [18,19] have shown that the simplified models by Volkersen and Goland-Reissner can be used to study the failure behavior of single lap joints but not all effects are covered to the full extent. In this work an extended weak interface model is used to model the adhesive joint, the model proposed by Ojalvo and Eidinoff [14]. The distinctive feature of this model is an extended consideration of the adhesive layer thickness effect and a linear distribution of the shear stresses in the adhesive layer. Let us consider a single lap joint under axial loading Fas shown in Fig. 2.. The height of the adherends shall be h, the thickness of the adhesive layer t and the length of the overlap shall beL. The width of the joint is denoted as b. Young's modulus of the adherend ist denoted as xE and Young's modulus of the adhesive as aE . The respective Poisson's ratios are denoted as and a. The shear modulus of the adhesive layer is denoted by aG .The axial coordinate x runs from the middle of the adhesive layer. Due to the large deformations of the adherends due to bending a non-linear dependence of the bending moment at the end of the overlap to the axial force Fis used: ( ) 2 h t M k F F (4) with ( ) k F being the non-linear moment factor. In literature many approaches can be found for defining this moment factor. In this work the moment factor given by Tsai and Morton [16] is used: 2 1 3 1 1 2 2 tanh 2 2 x k L F h E hb (5) The peel stress in the midplane (index 0) of the adhesive layer is obtained as follows [14]: 0 1 1 2 2 1 2 2 2 2 2 sinh sin cosh cos F x x x x A A bL L L L L (6) with 4 2 2 1,2 3 3 3 1 , . 2 2 2 4 a a x x E L G L t h E th E th (7) The constants 1A and 2A have to be obtained from the following boundary conditions. 2 3 2 0 0 3 2 2 2 3 4 3 1 a L L x x x F t E L L t k x h x E th b h (8)
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